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For the given graph, estimate π prime of two.
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Letβs begin by recalling exactly what is meant by this notation, π prime of two.
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It means the first derivative of our function π evaluated at the point where π₯ is equal to two.
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The first derivative of the function also gives the slope of its graph.
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So, by being asked to find the first derivative of π when π₯ equals two, weβre also being asked to find the slope of the graph of π at the point where π₯ equals two.
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Now, we can see from the figure that weβve been given that a tangent line has been drawn into the graph of π.
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Thatβs the graph in blue at the point where π₯ equals two.
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We can use this tangent line to approximate or estimate the slope of the graph itself at this point.
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Now, we know that to find the slope of a straight line which passes through the points with coordinates π₯ one, π¦ one and π₯ two, π¦ two, we can use the formula π¦ two minus π¦ one over π₯ two minus π₯ one, which we sometimes think of as change in π¦ over change in π₯ or rise over run.
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We, therefore, need to find the coordinates of two points that lie on this tangent line.
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We could use the point where π₯ equals two itself or we could use the points now marked in pink.
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Any two of these three points would be absolutely fine.
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It makes sense to choose points whose coordinates we can determine easily.
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So, I have chosen points which are located where grid lines intersect.
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The two points that Iβve chosen have coordinates 1.5, two and 2.5, negative six.
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If we look at the π₯-axis, we can see that each of the grid lines represents 0.5 units.
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So, substituting the coordinates of these points then, the slope of our line will be equal to negative six minus two over 2.5 minus 1.5.
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And this will give an approximation for the value of π prime of two.
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Simplifying, we have negative six minus two in the numerator, which is negative eight, and 2.5 minus 1.5 in the denominator, which is one.
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Negative eight over one is simply negative eight.
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And so, we have our approximation for π prime of two.
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Remember the key principles we used in this question were that the first derivative of our function gives the slope of its graph.
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And we can approximate the first derivative at any given point by finding the slope of the tangent to the curve at that point, which we can do using the standard formula for the slope of a straight line: π¦ two minus π¦ one over π₯ two minus π₯ one or change in π¦ over change in π₯.